Observe that the collection $\lbrace \be_n(\theta)\rbrace_{n\in\bZ}$ is an orthonormal basis of $L^2(\bR/2\pi\bZ)$. $\newcommand{\bT}{\mathbb{T}}$ For any positive integer $N$ we denote by $\bT^N$ the $N$-dimensional torus

where $(c_n)_{n\in\bZ}$ are i.i.d. Gaussian random variables with mean zero and variance $1$. $\newcommand{\eE}{\mathscr{E}}$ The correlation kernel of this random function is $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\vfi}{\varphi}$

Thus, the matrix $S^\ve(\vec{t})$ has a kernel if and only if the truncated characters $\chi^\ve_{t_1},\dotsc, \chi^\ve_{t_N}$ are linearly dependent. We show that this is not possible if $\vec{t}$ is nondegenerate and $\ve$ is sufficiently small.

Fix $\ve_0=\ve_0(N,w)$ such that if $\ve<\ve_0$ the support of $x\mapsto w(\ve x)$ contains a long interval of the form $[\nu_\ve, \nu_\ve+N-1]$, for some integer $\nu_\ve>0$. (Recall that $w$ is even.) In other words $\nu_\ve,\nu_\ve+1,\cdots,\nu_\ve+N-1\in Z_\ve$.

Let $\ve\in (0,\ve_0)$ and suppose that $\vez\in\bC^N\setminus 0$ and $\vec{t}\in\bT^N$ are such that such that $T_{\vez,\vec{t}}=0$. Thus

For $\vec{t}\in\bT^N$ nondegenerate we denote by $X(\vec{t})\subset H_\ve$ the vector space spanned by the characters $\chi_{t_j}$, $j=1,\dotsc, N$. $\newcommand{\bsD}{\boldsymbol{D}}$ Denote by $\bsD_N\subset H_\ve$ the space spanned by $\delta_1,\dotsc, \delta_n$. For $k=1,\dotsc, N$ we set $\newcommand{\bde}{\check{\delta}}$